There are 10 solid-colored balls in a box, including 1 green ball and

Question

There are 10 solid-colored balls in a box, including 1 green ball and 1 yellow ball. If 3 of the balls in the box are to be chosen at random, without replacement, what is the probability that the 3 balls chosen will include the green ball but not the yellow ball?

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

PS00896

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H

Bunuel · Accepted Answer

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

Approach #1:

We need to find the probability of GXX.  P(GXX)=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30} , we are multiplying by  \frac{3!}{2!}  since GXX scenario could occur in 3 ways: GXX, XGX, or XXG (the number of permutations of 3 letters GXX out of which 2 X's are identical).

Answer: B.

Approach #2:

P=\frac{C^2_8*C^1_1}{C^3_{10}}=\frac{7}{30} , where  C^2_8  is ways to select 2 other color balls out of 8,  C^1_1  is ways to select 1 green ball, and  C^3_{10}  is total ways to select 3 balls out of 10.

Answer: B.

Hope it's clear.

LalaB · Answer

the worst case is

Some color-some another color- Green

ok, we have 10 balls , 1 of them is yellow, and the another one is green.
need to find that worst case

3*(8/10)*(7/9)*(1/8)=7/30

here the integer 3 means that u can order Some color-some another color-green in 3 ways (green -some c-some an.c.; some c-green-some an.c.-some c.-some an.c.-green)

8/10 means that u can select 8 (10 minus 1 yellow minus 1 green color) out of 10 colors
7/9 means that u can select 7 (9 minus 1 yellow minus 1 green color) out of 9 remaining colors
1/8 means that u have only one green out of 8 remaining colors

hope it helps

EvaJager · Answer

Probability approach, direct not for the complementary event (it is more complicated, not worth trying in this case: it can be one G and also one Y or neither G nor Y, plus an additional different color):
P(G) * P(noG & noY) * P( another noG & noY) * 3 = 1/10 * 8/9 * 7/8 * 3 = 7/30.
We need the factor of 3 because the Green ball can be either the first, second or third.

Answer: B

imhimanshu · Answer

Hi Bunuel,

Thanks for the reply. I was able to solve it using Approach # 2. But I have doubt in Approach # 1 . Request you to please provide your comments.

Isn't it an assumption that remaining 8 balls are of same color. If the remaining 8 balls each of them are of different color, then the number of arrangements will get changed and hence probability.

Please tell me what I am missing
Thanks.
H

Bunuel · Answer

We are not assuming that 8 other balls are necessarily of some one particular color: they can be of one color or 8 different colors, but for us the only thing which is important that they are of different color than green and yellow.

Consider this: if the first ball selected is green then we are left with 9 balls out of which 1 is yellow and 8 other balls are not yellow (that's the only thing we care), so the probability of selecting non-yellow ball is 8/9.

Hope it's clear.

canada07 · Answer

Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

Bunuel · Answer

The probability of each case will be the same:
{GXX} -  P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90} ;

{XGX} -  P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90} ;

{XXG} -  P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90} ;

The sum:  \frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30} .

Hope it helps.

ScottTargetTestPrep · Answer

The number of ways to select 3 balls that include the green ball but not the yellow ball is:

1C1 x 1C0 x 8C2 = 1 x 1 x (8 x 7)/2! = 28 ways

(Note: 1C1 is the number of ways to select 1 green ball from the 1 green ball, 1C0 is the number of ways to select 0 yellow ball from the 1 yellow ball, and 8C2 is the number of ways to select 2 balls from the 8 remaining balls.)

The number of ways to select 3 balls from 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 120

Thus, the probability that the 3 balls chosen will include the Green ball but not the yellow ball is 28/120 = 7/30.

Alternate Solution:

Let’s first determine the probability of selecting the one green ball on the first draw and any of the remaining balls (except the yellow ball) on the succeeding two draws: 1/10 x 8/9 x 7/8 = 56/720.

Now, we see that we could also pick the green ball on the second draw, with any ball (except the yellow one) on the first and third draws, with probability: 8/10 x 1/9 x ⅞ = 56/720.

Similarly, we could instead pick the green ball on the third draw, with any ball (except the yellow one) on the first and second draws, with probability: 8/10 x 7/9 x 1/8 = 56/720.

Thus the total probability for the event where the green ball is drawn but the yellow one is not is:
56/720 + 56/720 + 56/720 = 168/720 = 21/90 = 7/30.

Answer: B

adityaganjoo · Answer

Bunuel  Need your help once more!

If we are choosing the balls without replacement, help me understand what is wrong with my approach:

3 possibilities: GXX, XGX, XXG

Probability = (1/10)(8/9)(7/8) + (9/10)(1/9)(7/8) + (9/10)(8/9)(1/8)
 = (1/720) 
 = (1/720) 
 = 191/720

Bunuel · Answer

Red figures are not correct.

Should be:
{GXX} -  P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90} ;

{XGX} -  P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90} ;

{XXG} -  P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90} ;

The sum:  \frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30} .

Thib33600 · Answer

How are you able to do the frst approach calculation without calculator?

Bunuel · Answer

\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30}

You don't need a calculator to get the value of the above. The key is to reduce.

8 gets reduced and we get:

\frac{1}{10}*\frac{1}{9}*7*\frac{3!}{2!}

\frac{3!}{2!}  is just 3, so we get:

\frac{1}{10}*\frac{1}{9}*7*3

That 3 also gets reduced to give:

\frac{1}{10}*\frac{1}{3}*7

And finally, we arrive at:

\frac{7}{30}

Hope it helps.

mcelroytutoring · Answer

Another tricky one!

We have three possible orders: 1) G/NY/NY 2) NY/G/NY, and 3) NY/NY/G. If we pick the "not yellow" balls before the green ball, then "not yellow" really means "not yellow OR green," because we have to save the green ball for later.

For example, in the order NY/NY/G, the first NY cannot be 9/10, because that includes the green ball, which we have to save for later. Thus, it must be 8/10 instead.

The same goes for the second NY, which must be 7/9.

https://www.drivecms.com/uploads/mcelroytutoring.com/Screen%20Shot%202024-02-01%20at%206.17.28%20PM.png

gmathero911 · Answer

Hi  Bunuel

Could you explain why it's (1/10)*(8/9)*(7/8)*(3!/2!) and not (1/10)*(8/9)*(7/8)*(10!/7!3!) ? Thank you!

Bunuel · Answer

3!/2! = 3 there represents the number of ways scenario (Green, Not Yellow, Not Yellow) can occur:

{Green, Not Yellow, Not Yellow}
{Not Yellow, Green, Not Yellow}
{Not Yellow, Not Yellow, Green}

Thus, 3!/2! is essentially the number of permutations of GXX.

DanTheGMATMan · Answer

Classic probability. We love it:

https://www.youtube.com/watch?v=wnPUxxNpoZo

egmat · Answer

This is a classic probability problem with constraints, and the key is understanding exactly what needs to happen. Let me walk you through this step-by-step.

Here's the core challenge you're facing:  You need to find the probability that when you randomly select 3 balls from 10, you MUST get the green ball and you MUST NOT get the yellow ball.

Let's break this down systematically:

Step 1: Calculate total possible outcomes

First, how many ways can you choose 3 balls from 10 total balls? This is a combinations problem since order doesn't matter.

Total outcomes =  C(10,3) = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = \frac{720}{6} = 120

So there are 120 different ways to select 3 balls from the box.

Step 2: Identify what makes an outcome "favorable"

Here's where you need to be really careful. For our selection to meet the requirements:
 
 The green ball MUST be included (so that's already 1 of our 3 balls) 
 The yellow ball MUST NOT be included (so we can't pick from this one) 
 We need 2 more balls to complete our set of 3

Notice what this means: We've used up the green ball and excluded the yellow ball. That leaves us with  10 - 1 - 1 = 8  remaining balls to choose from.

Step 3: Calculate favorable outcomes

Now you need to count how many ways you can pick 2 balls from these 8 remaining balls (the ones that are neither green nor yellow).

Favorable outcomes =  C(8,2) = \frac{8 	imes 7}{2 	imes 1} = \frac{56}{2} = 28

Step 4: Apply the probability formula

Probability =  \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{28}{120}

Simplify by dividing both numerator and denominator by 4:

\frac{28}{120} = \frac{7}{30}

Answer: B ( \frac{7}{30} )

The key insight:  Once you lock in the green ball (must include) and exclude the yellow ball (must not include), you're really just choosing 2 balls from the 8 that remain. This transforms a seemingly complex constraint problem into a straightforward combinations calculation.

---

You can check out the  step-by-step solution on Neuron by e-GMAT  to master the systematic approach for probability problems with constraints. The complete solution also covers the common mistakes students make (like miscounting the remaining balls as 9 instead of 8) and shows you the pattern recognition framework that applies to all constraint-based probability questions. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here .

Hope this helps!

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